A conveyor belt is moving at a constant speed | vedclass

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(A) The box is dropped on the moving belt,so its initial velocity relative to the belt is $u_{rel} = 2\, m/s$.The frictional force acting on the box p

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(A) The box is dropped on the moving belt,so its initial velocity relative to the belt is $u_{rel} = 2\, m/s$.The frictional force acting on the box provides an acceleration $a = \mu g = 0.5 	imes 10 = 5\, m/s^2$.This acceleration acts in the direction of the belt's motion to bring the box to rest relative to the belt.Using the kinematic equation $v_{rel}^2 = u_{rel}^2 - 2 a s$,where $v_{rel} = 0$ (when the box stops relative to the belt):$0 = (2)^2 - 2(5)s$$10s = 4$$s = 0.4\, m$.

$A$ conveyor belt is moving at a constant speed of $2\, m/s$. $A$ box is gently dropped on it. The coefficient of friction between them is $\mu = 0.5$. The distance that the box will move relative to the belt before coming to rest on it,taking $g = 10\, m/s^2$,is ........ $m$.

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